Exercise 1.3.5. Given the following system of equations:

find a solution to the system. (The exercise also calls for doing a sketch of the lines corresponding to the equations, as well as the line corresponding to the second equation after elimination. I’m skipping that part.)

Answer: We subtract 3 times the first equation from the second equation:

We can then back-substitute to solve for and then :

So the solution is , .

The first equation corresponds to the line at a 45 degree angle to the axis and intersecting the origin. The second equation corresponds to the line with slope and intersecting the axis at the point (0, 3). After elimination the second equation (or ) corresponds to a horizontal line intersecting the axis at the point (0, 2).

UPDATE: Corrected the -intercept for the second equation; thanks go to pana8a7 for finding and fixing this error. Also corrected a type in the elimination steps (had 19 instead of 18). Finally, took the time to convert all variable names and in-line equations to use LaTeX.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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3x+6y=18

3(x+2y)=18

x+2y=6

y=-x/2+3 not y=-x/2+6

You are correct. Thank you for finding and fixing this error! I have updated the post (and found and fixed another error at the same time).